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(C) The point $(\alpha, \beta)$ lies on the ellipse $25x^{2} + 4y^{2} = 1$,so we can write $\alpha = \frac{1}{5} \cos 	heta$ and $\beta = \frac{1}{2}

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$2929$

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(C) The point $(\alpha, \beta)$ lies on the ellipse $25x^{2} + 4y^{2} = 1$,so we can write $\alpha = \frac{1}{5} \cos 	heta$ and $\beta = \frac{1}{2} \sin 	heta$.The equation of a tangent to the parabola $y^{2} = 4x$ with slope $m$ is $y = mx + \frac{1}{m}$.Since it passes through $(\alpha, \beta)$,we have $\beta = m\alpha + \frac{1}{m}$,which simplifies to $m^{2}\alpha - m\beta + 1 = 0$.Let the slopes be $m_{1}$ and $m_{2}$ such that $m_{1} = 4m_{2}$.From the quadratic equation $m^{2}\alpha - m\beta + 1 = 0$,we have $m_{1} + m_{2} = \frac{\beta}{\alpha}$ and $m_{1}m_{2} = \frac{1}{\alpha}$.Substituting $m_{1} = 4m_{2}$,we get $5m_{2} = \frac{\beta}{\alpha}$ and $4m_{2}^{2} = \frac{1}{\alpha}$.Thus,$m_{2} = \frac{\beta}{5\alpha}$,so $4(\frac{\beta}{5\alpha})^{2} = \frac{1}{\alpha}$,which gives $4\beta^{2} = 25\alpha$.Substituting $\alpha = \frac{1}{5} \cos 	heta$ and $\beta^{2} = \frac{1}{4} \sin^{2} 	heta$,we get $4(\frac{1}{4} \sin^{2} 	heta) = 25(\frac{1}{5} \cos 	heta)$,so $\sin^{2} 	heta = 5 \cos 	heta$.$1 - \cos^{2} 	heta = 5 \cos 	heta \Rightarrow \cos^{2} 	heta + 5 \cos 	heta - 1 = 0$.Solving for $\cos 	heta$,we get $\cos 	heta = \frac{-5 \pm \sqrt{25 + 4}}{2} = \frac{-5 \pm \sqrt{29}}{2}$.Then $10\alpha = 2 \cos 	heta = -5 \pm \sqrt{29}$,so $(10\alpha + 5)^{2} = 29$.Also,$16\beta^{2} = 16(\frac{1}{4} \sin^{2} 	heta) = 4(1 - \cos^{2} 	heta) = 4(1 - (1 - 5 \cos 	heta)) = 20 \cos 	heta = 20(\frac{-5 \pm \sqrt{29}}{2}) = -50 \pm 10\sqrt{29}$.Thus,$(16\beta^{2} + 50)^{2} = (\pm 10\sqrt{29})^{2} = 100 	imes 29 = 2900$.Finally,$(10\alpha + 5)^{2} + (16\beta^{2} + 50)^{2} = 29 + 2900 = 2929$.

Column $I$	Column $II$
$(A)$ Circle	$(p)$ The locus of the point $(h, k)$ for which the line $h x+k y=1$ touches the circle $x^2+y^2=4$
$(B)$ Parabola	$(q)$ Points $z$ in the complex plane satisfying $\|z+2\|-\|z-2\|= \pm 3$
$(C)$ Ellipse	$(r)$ Points of the conic have parametric representation $x=\sqrt{3}\left(\frac{1-t^2}{1+t^2}\right), y=\frac{2 t}{1+t^2}$
$(D)$ Hyperbola	$(s)$ The eccentricity of the conic lies in the interval $1 \leq x < \infty$
	$(t)$ Points $z$ in the complex plane satisfying $\operatorname{Re}(z+1)^2=\|z\|^2+1$

If two tangents drawn from a point $(\alpha, \beta)$ lying on the ellipse $25x^{2} + 4y^{2} = 1$ to the parabola $y^{2} = 4x$ are such that the slope of one tangent is four times the other,then the value of $(10\alpha + 5)^{2} + (16\beta^{2} + 50)^{2}$ equals

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